Math 3 Quiz: Using Z Scores
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Using Z ScoresQuestion 1 of 20

An electronics manufacturer produces resistors with resistance values normally distributed around 100 ohms with standard deviation 5 ohms. A batch contains resistors measuring 92, 103, and 108 ohms. If an engineer needs to select one resistor that represents the most extreme deviation from specification, which should be chosen and why?

The 92-ohm resistor should be chosen because its z-score of -1.6 represents the largest negative deviation
The 108-ohm resistor should be chosen because its z-score of 1.6 represents the largest positive deviation
Either the 92-ohm or 108-ohm resistor because both have z-score magnitude 1.6, equally extreme
The 103-ohm resistor should be avoided, leaving either 92-ohm or 108-ohm as acceptable extreme examples
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Math 3 Quiz

Math 3 Quiz: Using Z Scores

Practice Using Z Scores in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Z Scores, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An electronics manufacturer produces resistors with resistance values normally distributed around 100 ohms with standard deviation 5 ohms. A batch contains resistors measuring 92, 103, and 108 ohms. If an engineer needs to select one resistor that represents the most extreme deviation from specification, which should be chosen and why?

  1. The 92-ohm resistor should be chosen because its z-score of -1.6 represents the largest negative deviation
  2. The 108-ohm resistor should be chosen because its z-score of 1.6 represents the largest positive deviation
  3. Either the 92-ohm or 108-ohm resistor because both have z-score magnitude 1.6, equally extreme (correct answer)
  4. The 103-ohm resistor should be avoided, leaving either 92-ohm or 108-ohm as acceptable extreme examples
Explanation: Calculate z-scores: 92Ω: z = (92-100)/5 = -1.6. 103Ω: z = (103-100)/5 = 0.6. 108Ω: z = (108-100)/5 = 1.6. For most extreme deviation, we want largest |z|. Both 92Ω and 108Ω have |z| = 1.6, making them equally extreme. Choice A and B each identify one extreme but miss that they're equally extreme. Choice D misunderstands the question's requirement.

Question 2

In a psychology experiment, reaction times are normally distributed with mean 800 ms and standard deviation 120 ms. A researcher observes that Participant 1's z-score is exactly twice that of Participant 2's z-score. If Participant 2 had a reaction time of 920 ms, what was Participant 1's reaction time?

  1. 1040 ms, because doubling Participant 2's z-score of 1.0 gives z = 2.0 for Participant 1 (correct answer)
  2. 1160 ms, because Participant 1's z-score of 2.0 corresponds to this reaction time value
  3. 680 ms, because Participant 1's z-score of 2.0 in the negative direction gives this time
  4. 560 ms, because doubling the deviation from the mean in the opposite direction yields this result
Explanation: First find Participant 2's z-score: z₂ = (920-800)/120 = 120/120 = 1.0. Participant 1's z-score is twice this: z₁ = 2(1.0) = 2.0. Convert back to reaction time: time = 800 + 2.0(120) = 800 + 240 = 1040 ms. Choice B has an incorrect time calculation. Choice C incorrectly assumes the doubled z-score should be negative. Choice D makes an unjustified assumption about opposite direction.

Question 3

Two athletes compete in different sports with normally distributed performance metrics. Athlete X has a z-score of 1.8 in swimming, while Athlete Y has a z-score of -2.1 in track. A sports analyst wants to determine which athlete's performance is more extreme relative to their respective sport's typical performance. What should the analyst conclude?

  1. Athlete X is more extreme because positive z-scores indicate above-average performance while negative z-scores show poor performance
  2. Athlete Y is more extreme because |-2.1| = 2.1 > |1.8| = 1.8, indicating greater distance from the mean (correct answer)
  3. Athlete X is more extreme because swimming requires more skill than track, making positive deviations more significant
  4. The athletes are equally extreme because both z-scores represent significant deviations in their respective normal distributions
Explanation: To determine which performance is more extreme, we compare the absolute values of the z-scores. |1.8| = 1.8 for Athlete X and |-2.1| = 2.1 for Athlete Y. Since 2.1 > 1.8, Athlete Y's performance is more extreme (further from the mean). Choice A incorrectly focuses on the sign rather than magnitude. Choice C introduces irrelevant information about sport difficulty. Choice D incorrectly claims they're equally extreme.

Question 4

A machine produces bolts with lengths normally distributed around 5.0 cm with standard deviation 0.2 cm. Quality control requires rejecting bolts with z-scores beyond ±2.0. If a bolt measures 5.35 cm and another measures 4.58 cm, what should quality control do?

  1. Accept both bolts because the first has z-score 1.75 and the second has z-score -2.1, with only one exceeding limits
  2. Reject both bolts because the first has z-score 1.75 and the second has z-score -2.1, with both outside acceptable range
  3. Accept both bolts because the first has z-score 1.75 and the second has z-score -1.9, with both within limits
  4. Reject the second bolt only because it has z-score -2.1 while the first has acceptable z-score 1.75 (correct answer)
Explanation: Calculate z-scores: First bolt: z = (5.35-5.0)/0.2 = 0.35/0.2 = 1.75. Second bolt: z = (4.58-5.0)/0.2 = -0.42/0.2 = -2.1. Since |1.75| < 2.0 but |-2.1| = 2.1 > 2.0, only the second bolt should be rejected. Choice A incorrectly suggests accepting both when one exceeds limits. Choice B incorrectly rejects the acceptable first bolt. Choice C has the wrong z-score calculation for the second bolt.

Question 5

In a manufacturing process, the weights of widgets are normally distributed with mean 250 grams and standard deviation 15 grams. Two widgets are randomly selected: Widget A weighs 280 grams and Widget B weighs 265 grams. A quality control inspector claims that Widget A is "more unusual" than Widget B. Which statement best evaluates this claim?

  1. The claim is correct because Widget A has a z-score of 2.0 while Widget B has a z-score of 1.0, and |2.0| > |1.0| (correct answer)
  2. The claim is incorrect because Widget B has a z-score of 1.5 while Widget A has a z-score of 1.0, and |1.5| > |1.0|
  3. The claim is correct because Widget A weighs 280 grams while Widget B weighs 265 grams, and 280 > 265
  4. The claim is incorrect because both widgets have the same z-score magnitude of 1.0, making them equally unusual
Explanation: To determine which widget is more unusual, we calculate z-scores. For Widget A: z = (280-250)/15 = 30/15 = 2.0. For Widget B: z = (265-250)/15 = 15/15 = 1.0. Since |2.0| > |1.0|, Widget A is indeed more unusual. Choice B has incorrect z-score calculations. Choice C incorrectly uses raw weights instead of standardized scores. Choice D incorrectly states both have the same z-score magnitude.

Question 6

A standardized test has scores normally distributed with mean 500 and standard deviation 100. Student A scored 650, Student B scored 720, and Student C scored 380. If a scholarship requires applicants to be in the top 2.5% of test takers (z-score ≥ 1.96), which students qualify?

  1. Only Student B qualifies with z-score 2.2, while Students A and C have z-scores 1.5 and -1.2 respectively (correct answer)
  2. Students A and B both qualify with z-scores 1.5 and 2.2, while Student C has z-score -1.2
  3. All three students qualify because Students A, B, and C have z-scores 1.5, 2.2, and 1.2 respectively
  4. No students qualify because the z-score requirement of 1.96 is too stringent for these particular test scores
Explanation: Calculate z-scores: Student A: z = (650-500)/100 = 1.5. Student B: z = (720-500)/100 = 2.2. Student C: z = (380-500)/100 = -1.2. Only Student B has z ≥ 1.96, so only B qualifies. Choice B incorrectly includes Student A whose z-score is below 1.96. Choice C incorrectly calculates Student C's z-score as positive. Choice D incorrectly concludes no one qualifies when Student B clearly does.

Question 7

A teacher curves exam scores using z-scores. The original scores are normally distributed with mean 76 and standard deviation 8. If a student scored 84 on the original exam, and the teacher adds 10 points to everyone's score, what is the student's z-score on the curved distribution?

  1. z = 2.0, because the curve adjustment amplifies the original deviation from the mean significantly
  2. z = 1.75, because the new score of 94 with mean 86 and unchanged SD gives this result
  3. z = 0.875, because adding 10 points changes both the numerator and denominator of the z-score calculation
  4. z = 1.0, because adding a constant to all scores doesn't change z-scores or standard deviation (correct answer)
Explanation: When you encounter z-score problems involving transformations, the key insight is understanding how different operations affect the mean, standard deviation, and ultimately the z-score itself. A z-score measures how many standard deviations a value is from the mean: z=x−μσz = \frac{x - \mu}{\sigma}. Let's trace what happens when the teacher adds 10 points to everyone's score. Originally: mean = 76, standard deviation = 8, student's score = 84 Student's original z-score: z=84−768=1.0z = \frac{84 - 76}{8} = 1.0 After adding 10 points to all scores:
  • New mean: 76 + 10 = 86
  • New standard deviation: still 8 (adding a constant doesn't change variability)
  • Student's new score: 84 + 10 = 94
New z-score: z=94−868=88=1.0z = \frac{94 - 86}{8} = \frac{8}{8} = 1.0 The z-score remains unchanged because adding a constant shifts all values equally, preserving their relative positions. Choice A incorrectly suggests the curve "amplifies" deviation—it doesn't. Choice B makes a calculation error, getting 1.75 instead of 1.0 when computing 94−868\frac{94-86}{8}. Choice C incorrectly assumes that adding a constant affects the standard deviation (the denominator), which it doesn't—only the relative spread matters. Study tip: Remember that z-scores are unaffected by adding or subtracting constants to all data points. Only operations that change the relative spread (like multiplication) will alter z-scores. This principle appears frequently in standardization problems.

Question 8

A factory produces items with weights normally distributed around 50 grams with standard deviation 8 grams. An inspector finds three items with weights 42g, 58g, and 66g. If the inspector can only investigate one item due to time constraints, which should be prioritized based on how unusual it is?

  1. The 42g item should be investigated because it has z-score -1.0, indicating it's underweight
  2. The 66g item should be investigated because it has z-score 2.0, the most extreme deviation (correct answer)
  3. The 58g item should be investigated because it has z-score 1.0, showing consistent overweight tendency
  4. Any item can be investigated since all three have z-scores within 2 standard deviations of normal
Explanation: Calculate z-scores: 42g: z = (42-50)/8 = -1.0. 58g: z = (58-50)/8 = 1.0. 66g: z = (66-50)/8 = 2.0. The most unusual item has the largest absolute z-score, which is the 66g item with |2.0| = 2.0. Choice A correctly calculates but focuses on direction rather than magnitude. Choice C correctly calculates but chooses a less extreme item. Choice D misses the point that among acceptable items, some are still more unusual.

Question 9

A pharmaceutical company tests reaction times for a new medication. Reaction times are normally distributed with mean 450 milliseconds and standard deviation 60 milliseconds. If a patient has a reaction time that corresponds to a z-score of -1.5, and another patient's reaction time is 30 milliseconds faster, what is the z-score for the second patient?

  1. z=−2.0z = -2.0 because the second patient's time is 30 milliseconds faster in a distribution with SD = 60 (correct answer)
  2. z=−1.0z = -1.0 because reducing the first patient's time by 30 milliseconds increases the z-score by 0.5
  3. z=−2.0z = -2.0 because the second patient's reaction time of 330 milliseconds converts to this z-score
  4. z=−1.0z = -1.0 because the second patient's reaction time of 360 milliseconds yields this standardized value
Explanation: First, find the first patient's reaction time: if z = -1.5, then time = 450 + (-1.5)(60) = 360 ms. The second patient is 30 ms faster, so their time is 360 - 30 = 330 ms. The z-score is (330-450)/60 = -120/60 = -2.0. Choice B incorrectly calculates the effect of the 30 ms change. Choice C has the right z-score but wrong reasoning about the time calculation. Choice D has incorrect time and z-score values.

Question 10

A medical study tracks blood pressure readings that are normally distributed with mean 120 mmHg and standard deviation 15 mmHg. Patient X has a reading with z-score 1.6. Patient Y has a reading that is 18 mmHg higher than Patient X. What can be concluded about Patient Y's z-score?

  1. Patient Y's z-score is 3.6
  2. Patient Y's z-score is 2.2
  3. Patient Y's z-score is 1.8
  4. Patient Y's z-score is 2.8 (correct answer)
Explanation: When you encounter z-score problems involving multiple patients or data points, remember that z-scores measure how many standard deviations away from the mean a value falls. The key insight is understanding how changes in raw values translate to changes in z-scores. First, let's find Patient X's blood pressure reading. Using the z-score formula: z=x−μσz = \frac{x - \mu}{\sigma}, we get 1.6=x−120151.6 = \frac{x - 120}{15}. Solving: x=120+(1.6)(15)=144x = 120 + (1.6)(15) = 144 mmHg. Patient Y's reading is 18 mmHg higher than Patient X's, so Patient Y has 144+18=162144 + 18 = 162 mmHg. Now we calculate Patient Y's z-score: zY=162−12015=4215=2.8z_Y = \frac{162 - 120}{15} = \frac{42}{15} = 2.8. Looking at the wrong answers: Choice A (3.6) incorrectly adds the z-scores: 1.6+1.8=3.41.6 + 1.8 = 3.4 (close to 3.6), but z-scores don't add this way. Choice B (2.2) might come from adding 1.6+0.61.6 + 0.6, perhaps confusing the 18 mmHg increase with a smaller z-score change. Choice C (1.8) represents just the change in z-score (1815=1.2\frac{18}{15} = 1.2, though this calculation is also flawed), not Patient Y's actual z-score. The correct answer is D: Patient Y's z-score is 2.8. Study tip: When dealing with z-score changes, always convert back to raw values first, make the specified change, then recalculate the new z-score. Don't try to directly manipulate z-scores unless you're certain about the mathematical relationship.

Question 11

Two normally distributed datasets have the same mean but different standard deviations. Dataset X has SD = 10, Dataset Y has SD = 5. If a value in Dataset X has z-score 1.2, what z-score would the same raw value have in Dataset Y?

  1. z = 0.6, because the z-score decreases proportionally when standard deviation decreases by half
  2. z = 2.4, because the z-score doubles when the standard deviation is halved (correct answer)
  3. z = 1.2, because z-scores are independent of standard deviation when means are equal
  4. z = 2.4, because smaller standard deviations make the same deviation from mean more significant
Explanation: If the value has z-score 1.2 in Dataset X, it's 1.2(10) = 12 units above the mean. In Dataset Y with the same mean but SD = 5, this same raw value would have z-score = 12/5 = 2.4. Choice A incorrectly reduces the z-score when SD decreases. Choice C incorrectly claims z-scores are independent of SD. Choice D has the right answer but incomplete reasoning about the relationship.

Question 12

A coffee shop tracks daily sales, which are normally distributed with mean $2400 and standard deviation $300. On Monday, sales were $2100. On Tuesday, sales increased by $450. How do the z-scores for these two days compare?

  1. Monday: z = -1.0, Tuesday: z = 0.5, so Tuesday's sales were more typical than Monday's
  2. Monday: z = -1.0, Tuesday: z = 0.5, so Monday's sales were more unusual than Tuesday's (correct answer)
  3. Monday: z = -1.0, Tuesday: z = 1.5, so Tuesday's sales were more unusual than Monday's
  4. Monday: z = -1.0, Tuesday: z = 1.5, so both days had equally unusual sales patterns
Explanation: Monday's z-score: z = (2100-2400)/300 = -300/300 = -1.0. Tuesday's sales: 2100 + 450 = 2550. Tuesday's z-score: z = (2550-2400)/300 = 150/300 = 0.5. Since |-1.0| = 1.0 > |0.5| = 0.5, Monday was more unusual. Choice A reaches the wrong conclusion despite correct calculations. Choice C incorrectly calculates Tuesday's z-score. Choice D has incorrect Tuesday z-score and wrong conclusion about equal unusualness.

Question 13

A quality control manager monitors production times that are normally distributed with mean 45 minutes and standard deviation 6 minutes. Three products have completion times of 39, 48, and 54 minutes. If the manager wants to identify which product's completion time is most typical, which should be selected?

  1. The 39-minute product because its z-score of -1.0 indicates it's exactly one standard deviation from mean
  2. The 48-minute product because its z-score of 0.5 is closest to zero, making it most typical (correct answer)
  3. The 54-minute product because its z-score of 1.5 shows it's within normal variation range
  4. The 48-minute product because its completion time of 48 minutes is closest to the mean of 45 minutes
Explanation: Calculate z-scores: 39 min: z = (39-45)/6 = -1.0. 48 min: z = (48-45)/6 = 0.5. 54 min: z = (54-45)/6 = 1.5. The most typical value has z-score closest to 0, which is the 48-minute product with z = 0.5. Choice A correctly calculates but misinterprets what 'typical' means. Choice C correctly calculates but chooses a less typical value. Choice D uses raw differences instead of standardized scores.

Question 14

In a large manufacturing company, employee salaries are normally distributed with a mean of $65,000 and a standard deviation of $12,000. If Sarah earns $77,000 and Miguel earns $59,000, which statement best describes their relative positions in the salary distribution?

  1. Sarah's salary is exactly 1 standard deviation above the mean, while Miguel's salary is exactly 0.5 standard deviations below the mean (correct answer)
  2. Sarah's salary is exactly 1 standard deviation above the mean, while Miguel's salary is exactly 1 standard deviation below the mean
  3. Sarah's salary is 1.2 standard deviations above the mean, while Miguel's salary is 0.5 standard deviations below the mean
  4. Sarah's salary is 0.8 standard deviations above the mean, while Miguel's salary is 0.6 standard deviations below the mean
Explanation: To find z-scores: For Sarah: z = (77,000 - 65,000) / 12,000 = 12,000 / 12,000 = 1.0. For Miguel: z = (59,000 - 65,000) / 12,000 = -6,000 / 12,000 = -0.5. Choice B incorrectly calculates Miguel's z-score as -1. Choice C uses incorrect arithmetic. Choice D makes calculation errors for both employees.

Question 15

A standardized test has scores that are normally distributed with a mean of 500 and a standard deviation of 100. If a student's z-score is -1.5, and another student scored 650, what is the difference between their raw scores?

  1. 300 points (correct answer)
  2. 250 points
  3. 200 points
  4. 150 points
Explanation: First student's raw score: Using z = (x - μ)/σ, we get -1.5 = (x - 500)/100, so x = 500 + (-1.5)(100) = 350. Second student scored 650. The difference is 650 - 350 = 300 points. Choice B results from miscalculating the first student's score as 400. Choice C represents just the z-score times standard deviation (1.5 × 100 + 50). Choice D uses only the z-score portion (1.5 × 100).

Question 16

Two different normal distributions are being compared. Distribution A has μ = 80 and σ = 15, while Distribution B has μ = 120 and σ = 20. A value of 95 from Distribution A and a value of 140 from Distribution B are being compared using z-scores. Which statement is correct?

  1. The value from Distribution A has a higher z-score and represents a more extreme position relative to its distribution
  2. The value from Distribution B has a higher z-score and represents a more extreme position relative to its distribution
  3. Both values have identical z-scores and represent equally extreme positions relative to their distributions (correct answer)
  4. The value from Distribution A has a lower z-score but represents a more extreme position relative to its distribution
Explanation: For Distribution A: z = (95 - 80)/15 = 15/15 = 1.0. For Distribution B: z = (140 - 120)/20 = 20/20 = 1.0. Both z-scores are identical at 1.0, meaning both values are exactly 1 standard deviation above their respective means. Choice A incorrectly assumes Distribution A has a higher z-score. Choice B incorrectly assumes Distribution B has a higher z-score. Choice D correctly identifies equal extremeness but incorrectly states Distribution A has a lower z-score.

Question 17

A researcher is comparing test performance across two different exams. Exam 1 has a mean of 78 and standard deviation of 12. Exam 2 has a mean of 85 and standard deviation of 15. If Maria scored 90 on Exam 1 and 94 on Exam 2, on which exam did she perform better relative to other test-takers?

  1. Both exams equally, because her z-scores were approximately the same on both exams
  2. Exam 2, because her z-score was approximately 1.0 compared to 0.6 on Exam 1
  3. Exam 2, because her raw score of 94 was higher than her raw score of 90 on Exam 1
  4. Exam 1, because her z-score was approximately 1.0 compared to 0.6 on Exam 2 (correct answer)
Explanation: When comparing performance across different exams with different means and standard deviations, you need to use z-scores to standardize the results. Raw scores alone can't tell you relative performance because the exams have different difficulty levels and score distributions. A z-score tells you how many standard deviations above or below the mean a score falls, using the formula: z=x−μσz = \frac{x - \mu}{\sigma} For Maria's Exam 1 performance: z=90−7812=1212=1.0z = \frac{90 - 78}{12} = \frac{12}{12} = 1.0 For Maria's Exam 2 performance: z=94−8515=915=0.6z = \frac{94 - 85}{15} = \frac{9}{15} = 0.6 Maria performed better relative to other test-takers on Exam 1, where she scored 1.0 standard deviation above the mean compared to only 0.6 standard deviations above the mean on Exam 2. Choice A is wrong because the z-scores aren't the same (1.0 vs 0.6). Choice B incorrectly assigns the z-scores to the wrong exams, swapping the values. Choice C falls into the trap of comparing raw scores directly—while 94 is higher than 90, this ignores that Exam 2 had a higher mean and more variability, making Maria's performance less impressive relative to her peers. Remember: when comparing performance across different tests or populations, always calculate z-scores first. Raw scores can be misleading when the underlying distributions differ. The higher z-score indicates better relative performance, regardless of which raw score is larger.

Question 18

A factory produces bolts with diameters that are normally distributed with mean 2.50 cm and standard deviation 0.08 cm. Quality control rejects bolts that are more than 2 standard deviations away from the mean in either direction. If a bolt measures 2.35 cm in diameter, what can be concluded about this bolt?

  1. The bolt should be rejected because its z-score of -1.875 exceeds 2 standard deviations below the mean
  2. The bolt should be accepted because its z-score of -1.875 is within 2 standard deviations of the mean (correct answer)
  3. The bolt should be rejected because its z-score of -2.125 exceeds 2 standard deviations below the mean
  4. The bolt should be accepted because its diameter of 2.35 cm is above the minimum acceptable diameter
Explanation: Calculate z-score: z = (2.35 - 2.50)/0.08 = -0.15/0.08 = -1.875. Since |-1.875| < 2, the bolt is within 2 standard deviations and should be accepted. Choice A correctly calculates the z-score but incorrectly concludes it exceeds 2 standard deviations. Choice C uses an incorrect z-score calculation. Choice D ignores the z-score criterion and uses an unstated minimum diameter.

Question 19

A quality control manager knows that the weights of products are normally distributed with a mean of 250 grams and a standard deviation of 18 grams. If a product weighs 232 grams, and the manager wants to find another product that is exactly as far above the mean as this product is below the mean, what should the second product weigh?

  1. 268 grams (correct answer)
  2. 286 grams
  3. 264 grams
  4. 272 grams
Explanation: First, find the z-score of the 232-gram product: z = (232 - 250)/18 = -18/18 = -1.0. A product equally far above the mean would have z = +1.0. Converting back: x = 250 + (1.0)(18) = 268 grams. Choice B (286) adds twice the difference to the mean (250 + 36). Choice C (264) represents half the correct distance above the mean. Choice D (272) uses incorrect arithmetic in the conversion.

Question 20

A standardized test is administered to two different populations. Population A has mean score 480 with standard deviation 60, while Population B has mean score 520 with standard deviation 40. A student from Population A scored 540, and a student from Population B scored 580. Which conclusion about their relative standings is most accurate?

  1. The Population A student performed better relative to their group, with a z-score of 1.0 versus 1.5 for the Population B student
  2. The Population B student performed better relative to their group, with a z-score of 1.5 versus 1.0 for the Population A student (correct answer)
  3. Both students performed equally well relative to their respective populations, both achieving z-scores of approximately 1.2
  4. The Population B student performed better because their raw score of 580 exceeds the Population A student's score of 540
Explanation: Population A student: z = (540-480)/60 = 60/60 = 1.0. Population B student: z = (580-520)/40 = 60/40 = 1.5. The higher z-score (1.5) indicates better relative performance. Choice A reverses the z-score assignments. Choice C incorrectly calculates both z-scores as 1.2. Choice D incorrectly compares raw scores instead of relative performance within each population.